ADVANCED PROBLEMS AND SOLUTIONS Edited by

RAYMOND E.WHITNEY Lock Haven State College, Lock Haven, Pennsylvania

Send all communications concerning Advanced Problems and Solutions to Raymond E. Whitney, Mathematics Department, Lock Haven State College, Lock Haven, Pennsylvania 17745. This department especially welcomes problems believed to be new or extending old resul ts . Proposers should submit solutions or other information that will ass is t the editor. To facilitate their consideration, solutions should be submitted on separate signed sheets within two months after publication of the problems.

H-195 Proposed by Verner £ Hoggatt, Jr., San Jose State College, San Jose, California

Consider the array indicated below:

1

1

2

5

13

34

89

1

2

4

9

22

56

145

1

3

7

16

38

1

4

11

27

65

1

5

16

1

6

22 1 1

(i) Show that the row sums are F„ , n ^ 2. (ii) Show that the rising diagonal sums are the convolution of

( F 2 n - l ^ = 0 a n d { ^ 2 , 2 ) } ; = 0 ,

the generalized numbers of Harr i s and Styles.

H-196 Proposed by J. B. Roberts, Reed College, Portland, Oregon.

(a) Let A0 be the set of integral par ts of the positive integral multiples of r, where

1 + ^ 5 T = „ —

413

414 ADVANCED PROBLEMS AND SOLUTIONS [Oct.

and let A - , m = 0, 1, 2, • • • , be the set of integral par t s of the numbers nr2

for n 61 A . Prove that the collection of Z of all positive integers is the disjoint union of the A..

J (b) Generalize the proposition in (a).

H-197 Proposed by Lawrence Somer, University of Hi in o is, Urbana, Illinois.

n sion relationship:

Let { i r } _ be the t-Fibonacci sequences with positive entries satisfying the recur -

Find

(t) u n

. l im n —•

t Vu . Z—i n- i i=l

(t) u n+1 u oo n

SOLUTIONS HYPER -TENSION

H-185 Proposed by L Carlitz, Duke University, Durham, North Carolina.

Show that

(1 - 2x)n = £ ( - D n " k ( n 2

+k

k ) ( 2k

k ) (1 - v ) n - k2 F l [-k; n + k + 1; k + 1; x]

k=0 ^ / \ /

where 2 F i [a ,b ; c; x] denotes the hypergeometric function.

Solution by the Proposer.

We star t with the identity

r s E (2r + 3s)! (y - z) z _

r!s!(r + 2s)! ^ + .2r+3s+l 1 - y - z r , s=0 y*

Now put y = u + v, z = v, so that

frv V ( 2 r + 3s)L u v = 1 1 ; JLJ r l s l ( r + 2s)l ,- + ,2r+3s+l 1 - u - 2v

r . s=0 * U v '

1972] ADVANCED PROBLEMS AND SOLUTIONS 415

The right-hand side of (*) is equal to

n=0

while the left-hand side

oo oo

E (2r + 3s)l r s V 1 / -x\kf2r + 3s + k \ , ^ ± r l s l ( r + 2s)l U V ^ {~1} { k J ( U + V)

r , s=0 k=0

It follows that

, , . ,n \ "^ , , ,k v ^ . n x k / 2 r + 3s + k \ (2r + 3s)l r s <u + 2 v > = L <u + v) L ('1] \ k J rl8l(r + 2s)l U v

k=0 r+s=n-k

E , i x n - r - s (r + 2s + n)l r s , , vn-r-s

W r '.si(r + 2s)l (n - r - s)l u v ( u + v ) r+s<n

n . k An-k E t l Nn-k (u + v) \7* (s + n + k)l k-s s

("1 } (n - k)l L* sl<k - s)l(k + s)i u v ' k=0 s=0

Taking u = 1, v = -x , we get

(1 - 2x)n = £ (-Dn"k EMV- 'MI (1 " V)n_k ^ ^ n + k + 1; k + 1; X] klkl(n - k)t

k=0

A CONGRUENCE IN ITS PRIME

H-186 Proposed by James Desmond, Florida State University, Tallahassee, Florida.

The generalized Fibonacci sequence is defined by the recurrence relation

u , + u = u ^ , n-1 n n+1

where n is an integer and U0 and U a re arbi t rary fixed integers. For a prime p and integers n, r , s and t, show that

u _, = U ^ (mod p) , np+r sp+t x ^ 3

416 ADVANCED PROBLEMS AND SOLUTIONS [Oct.

if p = ±1 (mod 5) and n + r = s + t, and that

U _,_ = ( - l ) r + t U ^ (mod p) np+r sp+t ^

if p = ±2 (mod 5) and n - r = s - t.

Solution by the froposer.

We have from Hoggatt and Buggies, "A P r i m e r for the Fibonacci Sequence — Par t III ," Fibonacci Quarterly, Vol. 1, No. 3, 1963, p. 65, and by Fe rma t ' s theorem, that

p F ^ = V* ( ? V-^ F1 FP"i = F FP .. + F _, FP = F F - + F _,_ F (mod p)

np+r L^d 1 1 1 l+r n n-1 r n-1 p+r n r n-1 p+r n v

i=0 x /

for all n and r. From I. D. Ruggles, "Some Fibonacci Results Using Fibonacci-Type Se-quences," Fibonacci Quarterly, Vol. 1, No. 2, 1963, p. 79, we have that

F .^ . = F. . F . + F .F . -1+3 l + l j I j -1

for all i and j . Therefore,

F _,_ = F F , + F ^ F F + F F . F (mod p) np+r r n-1 r+1 p n r p -1 n ^

for all n and r. We have from Hardy and Wright, Theory of Numbers, Oxford University P r e s s , London, 1954, p. 150, that

F _- = 0 (mod p) and F = 1 (mod p)

if p = ±1 (mod 5), and that

F + i - • ° (mod p) and F = 1 (mod p)

if p = ±2 (mod 5). Let p = ±1 (mod 5) and n + r = s + t. Then

F _,_ = F F n + F ^ n F E F ^ (mod p) np+r r n-1 r+1 n r+n v'

for all n and r. Therefore

F .* = F ^ = F ^ = F _,_ (mod p). sp+t s+t n+r np+r *

It is easily verified by induction that

1972] ADVANCED PROBLEMS AND SOLUTIONS 417

U = t L F + UnF n n I n 0 n-1

for all n. Therefore

U n p + r s U l F n p + r + U 0 F n p + r - l = U l Fs p + t + U 0 F

S p + t - l S U sp + t < m o d P> •

Now, let p = +2 (mod 5) and n - r = s - t. From page 77 of the reference to Ruggles, we have

F .^ . - F.L. = ( - l ) j + 1 F . . i+J 1 3 i-3

for all i and j . Therefore

F ^ = F F - - F , - F + F F E F F - + F ^- F - 2F , - F + F F np+r r n-1 r+1 n r n r n-1 r+1 n r+1 n r n = F ^ - L F = ( - l ) r + 1 F (mod p)

r+n r n ' n - r F

for all n and r . Thus

( - l ) r + t F _,. = ( - l ) r + t ( - l ) t + 1 F . = ( - l ) r + 1 F = F _,_ (mod p) . v sp+t \ / \ / s _ t n - r np+r v

Hence U n p + r = U l F n P + r + U 0 F n P + r - l B U l ( - 1 ) r + t F s P + t + V- l ) ' - 1 * - 1 F

sp+t-l B ( " 1 ) r + t ( U l F s P + t + U 0 F

s p + t - l > - < - 1 ) r + t u s P + t ( m o d P>"

FIBONACCI IS A SQUARE

H-187 Proposed by Ira Gessel, Harvard University, Cambridge, Massachusetts.

Problem: Show that a positive integer n is a Fibonacci number if and only if either 5n2 + 4 or 5n2 - 4 is a square.

Solution by the Proposer.

Let F0 = 0, F4 = 1, F - = F + F _- be the Fibonacci ser ies and L0 = 2, ^ = 1, L ,- = L + L n be the Lucas se r ies . It is well known that r+1 r r - 1

(1) (-I)" + F*r = F r + 1 F r _ 1

(2) L r = F r + 1 + F r _ x

Subtracting four times the first from the square of the second equation, we have

whence L2 _ 4 ( - l ) r - 4F2 = (F _,, - F - )2 = F2 , r r r+1 r - 1 r

5F2 + 4 ( - l ) r = L2 . r r

418 ADVANCED PROBLEMS AND SOLUTIONS [Oct.

Thus if n is a Fibonacci number, either 5n2 + 4 or 5n2 - 4 is a square. I have two proofs of the converse. F i r s t Proof. We use the theorem (Hardy and Wright, An Introduction to the Theory of

Numbers, p. 153) that if p and q are integers , x is a real number, and |(p/q) - x | < l /2q2 , then p /q is a convergent to the continued fraction for x, and that (Hardy and Wright, p. 148) the convergents to the continued fraction for (1 + *s/5)/2 in lowest t e rms are F - / F . r

Assume that 5n2 ± 4 - m2. Then since m and n have the same parity, k = (m + n)/2 is an integer. Then substituting m = 2k - n in 5n2 ± 4 = m2, we get k2 - kn - n2 = ±1, so that k and n are relatively pr ime and

±l /n2 = (k/n)2 - (k/n) - 1 = [(k/n) - (N/5 + l ) /2 ] [(k/n) + (\/5 - l ) /2 ] .

Thus

| (k/n) - (s/5 + l ) / 2 | = 1/n2 | (k/n) + (^5 - l ) / 2 | .

Since 1 is a Fibonacci number, we may assume n ^ 2. Then

(2k - n)2 = m2 ^ 5n2 - 4 = 4n2 + (n2 - 4) ^ 4n2 ,

so 2k - n ^ 2n, whence k/n 2= 3/2. Thus (k/n) + (\/"5 - l ) / 2 > 2, so by the two theorems quoted above, k/n = F - / F for some r , and since both fractions are reduced, n = F .

Second Proof. Assume 5n2 ± 4 = m2. Then m2 - 5n2 = ±4, so

m + \/5n m - N/5n _ n " 2 * 2 ±l '

and since m and n have the same parity,

m + \ r5n , m - V5n _ — _ and *

are integers in Q ( N / 5 ) , where Q is the rationals, and since their product is ±1, they are units. It is well known (Hardy and Wright, p. 221) that the only integral units of Q(\/I>) are of the form ±x~ , where x = (1 + \/"5 )/2.

Then we have

(m + ^ 5 n ) / 2 = x r = \ (xr + y r ) + f L ^ ^ L • N/5 , 2 L ^ 5 J

where y = -1/x . Now x. + y = L and

1972] ADVANCED PROBLEMS AND SOLUTIONS 419

(Xr - y r ) / ^ 5 = F r

(Hardy and Wright, p. 148). Thus

| ( m + N/5n) = | ( L r + ^ 5 F r > ,

so n = F .

SUM SERIES

H-189 Proposed by L Carlitz, Duke University, Durham, North Carolina (Corrected).

Show that

(2r + 3s)l (a - b y ) r b S y r + 2 s

^ r l s ! ( r + 2s)l H _, ,2r+3s+l , , 2

r 9 S = 0 ' (1 + ay) 1 - ay - by2

Solution by the Proposer.

Put

i - ar

so that

1 - ax - bx2 A

m=0

m x m

1 = E * (1 - ax - bx2)(l - y) n '

J m,n=0

v n i n x y

Replacing y by x" y this becomes

1 V ^ m ~ n n

• -z / J G x y m,n=0 (1 - ax - bx2)(l - x y)

Hence that par t of the expansion of

(1 - ax - bx2)(l - x V )

that is independent of x is equal to

420 ADVANCED PROBLEMS AND SOLUTIONS [Oct.

(*) 1 - ay - by2

On the other hand, since

(1 - ax - bx2)(l - x~ y) = (1 + ay) - x(a - by) - bx2 - x" y ,

we have

(1 - ax - b x ^ U - x - V ) " 1 = £ [ ^ a - b ^ + b x ^ + x - V l

vT, S t E (r + s + t)l (a - by) b y r+2s- t rlsltl ,- ^ x r + s + t + l x

r,s,t=0 ( 1 + a ^

The par t of this sum that is independent of x is obtained by taking t = r + 2s. We get

E (2r + 3s)l (a - b y ) V y r 2 s

rT.sI(r + 2s)l ,.. , v2r+3s+i r , s=0 ( 1 + a y )

Since this is equal to (*), we have proved the stated identity.

IT 'S A MOD WORLD

H-190 Proposed by H. H. Ferns, Victoria, British Columbia.

Prove the fo l lowing

2 r F = n (mod 5)

2 r L = 1 (mod 5) ,

where F and L are the n Fibonacci and n Lucas numbers , respectively, and r is the least residue of n - 1 (mod 4).

Solution by the Proposer.

In an unpublished paper by the proposer , it i s shown that

^\- E (2kn

+1) 5 k

k=0

1 9 7 2 ] ADVANCED PROBLEMS AND SOLUTIONS 421

Hence

IT] k=l x '

Thus

5 k .

2 n 1 F n = n (mod 5)

Let n - 1 = 4m + r , where 0 < r < 4. Then

But

Hence

2 F = 2 F = 2 2 F = n (mod 5) .

24m = (24}m ^ 1 ( m o d 5 ) ^

To prove

use

2 r F = n (mod 5) .

2 r L = 1 (mod 5) ,

M k=0 % '

(which is derived in the same paper) and proceed as above*

JUST SO MANY TWO'S

H-192 Proposed by Ronald Alter, University of Kentucky, Lexington, Kentucky.

I f 3n+l

c n

3=0

•M1+J.

. • s (%: JH prove that

c = 26 n + 3 . Ns (N odd, n >0) .

n

Solution by the Proposer.

In the sequence

bk = b k - l " 3 b k - 2 ' ( k " l j bi = b2 = D >

it is easy to show that 2 is the highest power of 2 that divides b. if and only if k = 3 (mod 6), Also, by deriving the appropriate Binet formula, it follows that

422 ADVANCED PROBLEMS AND SOLUTIONS [Oct.

Thus ^im-c-^ i - »-k ^ 1

2 j=o

The desired result follows by observing

6n+3 6n+2 n *

Editorial Note: Please submit solutions for any of the problem proposals. We need fresh blood!

A GOLDEN SECTION SEARCH PROBLEM

REXH.SHUDDE Waval Post Graduate School, Monterey, California

After tiring of using numerous quadratic functions as objective functions for examples in my mathematical programming course , I posed the following problem for myself: Design a unimodal function over the (0,1) interval which is concave, has a maximum in the interior of (0,1), and is not a quadratic function. The purpose was to demonstrate numerically the golden section search.*

My first thoughts were to add two functions which are concave over the (0,1) interval with the property that one goes to -°o at 0 and the other goes to -°o at 1. My two initial choices were log x and l / (x - 1). The golden section search s tar t s at the two points xt = 1 - (1/0) and x2 = l/(p where 0 = (1 + N / 5 ) / 2 . After searching with 8 points, I noticed that the interval of uncertainty still contained the first search point so I thought it about time to find the location of the maximum analytically. I was dumfounded to discover that if I contin-ued indefinitely with the search my interval of uncertainty would still contain the initial search point.

* Douglas J. Wilde, Optimum Seeking Methods, Prentice Hall, Inc. (1964).